A long solenoid carries a current I(t) = I0 1 e t , where I0 = 30 A and = 1.6 s1

Question

A long solenoid carries a current I(t) = I0 1 e t , where I0 = 30 A and = 1.6 s1 . Inside this solenoid, and coaxial with it, is a smaller coil. The permeability of free space is 1.25664 × 106 Wb/A.

Inside coil has 250 turns

Outside solenoid has 584 turns

What is the magnetic flux which passes through one of the turns of the smaller coil at t = 3 s? Answer in units of T m2 .

011 (part 2 of 4) 10.0 points What is the mutual inductance? Answer in units of H. 012 (part 3 of 4) 10.0 points What is the magnitude of the maximum emf induced in the coil? Answer in units of V

(part 4 of 4) 10.0 points If the axis of the coil in the figure is rotated an angle 24 from the horizontal, what is the new mutual inductance of the coil in the solenoid? Answer in units of H.

Explanation / Answer

1)

magnetic field due to solenoid B = uo*n*I = uo*N1/L*I

flux = B*A = uo*N1/L*I*pi*r^2

at time t = 3

flux = 1.25664*10^-6*(584/8)*30*(1-e^-(1.6*3))*pi*0.063^2

flux = 3.403*10^-5 T <<----------answer

(2)

total flux = N2*B*A

but toal flux = M*I

N2*B*A = M*I

N2*uo*(N1/L)*I*pi*r^2 = M*I

N2*uo*(N1/L)*pi*r^2 = M

M = 250*1.25664*10^-6*(584/8)*pi*0.063^2

M =0.000286 H

++++++++++++++

maximum EMF = rate of chage in flux

EMF = (d/dt)*(N2*uo*(N1/L)*I*pi*r^2)

EMF = N2*uo*(N1/L)*I0*e^-(alpha*t)*pi*r^2*alpha

EMF is maximum at t = infinity

EMF = 250*1.25664*10^-6*(584/8)*30*1*pi*0.063^2*1.6

EMF = 0.0137 v

_____________

M' = M*costheta

M' = 0.0002613 H

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.